SLIDE 15 UTA(DIS)-poly
UTA-poly - Example II
◮ Scores of a1, a2 and a3 are given by :
U(a1) = px,0 + 10px,1 + 100px,2 + 1000px,3 + py,0 + 7py,1 + 49py,2 + 343py,3, U(a2) = px,0 + 6px,1 + 36px,2 + 324px,3 + py,0 + 8py,1 + 64py,2 + 512py,3, U(a3) = px,0 + 7px,1 + 49px,2 + 343px,3 + py,0 + 5py,1 + 25py,2 + 125py,3.
◮ We have a1 ≻ a2 and a2 ≻ a3, which implies :
U(a1) − U(a2) + σ+(a1) − σ−(a1) − σ+(a2) + σ−(a2) > 0, U(a2) − U(a3) + σ+(a2) − σ−(a2) − σ+(a1) + σ−(a1) > 0.
◮ By replacing U(a1), U(a2) and U(a3), we have :
4px,1 + 64px,2 + 776px,3 − py,1 − 15py,2 − 231py,3 + σ+(a1) − σ−(a1) −σ+(a2) + σ−(a2) > 0, −px,1 − 13px,2 − 19px,3 + 3py,1 + 39py,2 + 387py,3 + σ+(a2) − σ−(a2) −σ+(a3) + σ−(a3) > 0.
University of Mons
- O. Sobrie - N. Gillis - V. Mousseau - M. Pirlot - July 5, 2016
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